Tough LCM Problem Solving

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Tough LCM Problem Solving

by khandelwal.ab » Sun Apr 22, 2012 4:02 am
Can any expert help me with an efficient solution and sound reasoning for this problem?

Q. An alarm is set in three watches for 6:00 AM. The alarms ring for 20, 50 and 60 seconds, respectively and then go off to snooze mode for 2, 5 and 6 minutes respectively. At what time will the alarms start ringing simultaneously for the second time?

A) 6:30 AM
B) 6:42 AM
C) 6:35 AM
D) 7:12 AM
E) 7:30 AM

[spoiler]
OA: C
[/spoiler]

Thanks in advance!

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by GMATGuruNY » Sun Apr 22, 2012 4:47 am
khandelwal.ab wrote:Can any expert help me with an efficient solution and sound reasoning for this problem?

Q. An alarm is set in three watches for 6:00 AM. The alarms ring for 20, 50 and 60 seconds, respectively and then go off to snooze mode for 2, 5 and 6 minutes respectively. At what time will the alarms start ringing simultaneously for the second time?

A) 6:30 AM
B) 6:42 AM
C) 6:35 AM
D) 7:12 AM
E) 7:30 AM

[spoiler]
OA: C
[/spoiler]

Thanks in advance!
The alarms first go off at 6am.
The cycle for the longest alarm = 60 seconds + 6 minute snooze = 7 minutes.
Thus, the time needed must be a multiple of 7.
The only viable answer choices are B (6:42 implies that 42 minutes are needed) and C (6:35 implies that 35 minutes are needed).

Since the earlier time is 6:35am, let's see whether the other two alarms will go off at this time.

Answer choice C: 35 minutes
The cycle for the first alarm = 20 seconds + 2 minute snooze = 7/3 minutes.
Total time/cycle time = 35/(7/3) = 15 cycles.
The cycle for the second alarm = 50 seconds + 5 minute snooze = 35/6 minutes.
Total time/cycle time = 35/(35/6) = 6 cycles.
Success!
In 35 minutes, the first alarm will complete 15 cycles, the second alarm will complete 6 cycles, and the third alarm will complete 5 cycles (since 35/7 = 5 cycles), implying that all 3 alarms will go off at 6:35am.

The correct answer is C.
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by khandelwal.ab » Sun Apr 22, 2012 5:09 am
Awesome! thanks for the amazing explanation Mitch...this makes complete sense.

The question's source's explanation:

LCM of (120+20), (300+50) and (360+60) =>2100 secs. => 35 minutes.

was not making much sense to me. But I can understand it now..Thanks a lot!! :D

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by Shalabh's Quants » Mon Apr 23, 2012 4:15 am
khandelwal.ab wrote:Awesome! thanks for the amazing explanation Mitch...this makes complete sense.

The question's source's explanation:

LCM of (120+20), (300+50) and (360+60) =>2100 secs. => 35 minutes.

was not making much sense to me. But I can understand it now..Thanks a lot!! :D
I think question source's explanation is great! Lets see!

I clock will ring in 120+20 secs. second time = 140 sec.;

II clock will ring in 300+50 secs. second time = 350 sec.;

III clock will ring in 360+60 secs. second time = 420 sec.;

For all 3 clocks to ring simultaneously; take LCM of 140, 350 & 420. Which is 2100 sec or 35 min.
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by [email protected] » Tue Apr 24, 2012 4:29 pm
Yes, this question is very similar to the races question.

Let us say that 3 people starting from the same point are racing. A takes 2 mins to complete a lap, B takes 3 mins to complete 1 lap and C takes 1.5 mins. to complete a lap.

What is the time after which all the three will meet at the starting point again.

The answer would be the LCM of all the three timings given.

Another question could also have been something like, What is the time after which all the three will meet. (Not necessary the starting point).

In those cases, the normal time, distance and speed formula comes into picture, with relative speed concept.

However, in the current question, the snooze time and the alarm times are to be added is the point where I got stuck.

Otherwise the concept was taken from the races funda.

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